Write-Up #1


I chose to investigate Problem 2 from Assignment #1. The problem states to make up linear functions f(x) and g(x) and to investigate the following:
i. h(x)=f(x)+g(x)
ii. h(x)=f(x)*g(x)
iii. h(x)=f(x)/g(x)
iv. h(x)=f(g(x))
First, I began with very simple linear equations and I chose f(x)=x and g(x)=-x.

In this case, it is apparent that the graph of f(x), g(x), f(x)+g(x), and f(x)*g(x) all intersect at the origin (0,0).

Now, let's investigate what happens when we increase the complexity of the graph of f(x) just a little bit. Let f(x)=x+2 and g(x)=-x.

In this graph, the graphs of f(x), f(x)*g(x), and f(x)/g(x) intersect at the points (-2,0) and (-1,1).

What happens when f(x)=x+2 and g(x)=x-3?


In the third investigation, there are still at least 3 graphs that intersect. In this case, they are f(x), f(x)*g(x), and (f(x)/g(x). This matches up with the graphs that intesected in the second set of linear equations.

Let's dig deeper into the problem. What would happen if f(x) and g(x) are even more complex?

Let's let f(x)=5x-3 and g(x)=2x-5.


Once again the same three graphs of f(x), f(x)*g(x), and (f(x)/g(x) appear and intersect.

So from this investigation, it SEEMS that for arbitrary linear equations, f(x) and g(x), the graphs of f(x), f(x)g(x), and f(x)/g(x).

Though this is not even close to a proof, it is interesting to see that there is something that causes these three graphs to intersect regardless of the choice of linear equations.


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